POJ 1007 DNA Sorting

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最新推荐文章于 2021-08-13 22:12:51 发布

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DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 90543		Accepted: 36373

Description

One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be---exactly the reverse of sorted).

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length.

Input

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

Output

Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. Since two strings can be equally sorted, then output them according to the orginal order.

Sample Input

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

Source

East Central North America 1998

顺着POJ往下做的一道水题。用的是暴力的模拟方法。16MS。仔细想了下，可以优化的地方还蛮多的。

#include <stdio.h>
#include <algorithm>
using namespace std;
typedef struct
{
        char s[55];
        int cnt;
}Node;
Node a[105];
bool cmp(Node x,Node y)
{
      return x.cnt<y.cnt;
}
int main()
{
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
                scanf("%s",a[i].s),a[i].cnt=0;
                for(int j=0;j<n;j++)
                        for(int k=j+1;k<n;k++)
                                if(a[i].s[j]>a[i].s[k])
                                        a[i].cnt++;
        }
        sort(a,a+m,cmp);
        for(int i=0;i<m;i++)
                printf("%s\n",a[i].s);
        return 0;
}